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Grade 11Differential Calculus

Let f: R--> R be a function defined by f(x) = min{x + 1,|x|+1}; then prove that f(x) is diffrentiable everywhere

Profile image of Meghana
7 Years agoGrade 11
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2 Answers

Profile image of Arun
7 Years ago
Dear Meghana
 
This can be seen as problem if arise will be due to modulus function i.e. |x| + 1
 
Now it will be non differentiable at x= 0 but at x = 0 
 
x+ 1 will be taken as main function.
 
hence f(x) is diffrentiable everywhere
Profile image of Aditya Gupta
7 Years ago
the correct answer would be f(x)= x+1 (because |x| greater than equal to x for all x)
so f’(x)=1 and hence it is differentiable everywhere